Please help. [:
cosΘ = 0 ==> Θ = π/2, 3π/2
I think you've miscopied something in the question because the equation you've posted is a false statement. No "solution" is possible.
π/4 + 1 ≠ π/2 or 3π/2
...
It is a false statement. cos(π/4 + 1) is about –0.213. There is no variable to solve for
cos(a-b)=(cosa)(cosb)-(sina)(sinb)
pi/4=180/4=45 degrees
1=0 degrees
(cos45)(cos0)-(sin45)(sin0)
(sqrt2/2)(1)-(sqrt2/2)(0)
sqrt2/2
cos((π/4) + 1)= cos((π/4) = 45 degrees
you can't. cos (0.25pi + 1 ) is about -.2
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cosΘ = 0 ==> Θ = π/2, 3π/2
I think you've miscopied something in the question because the equation you've posted is a false statement. No "solution" is possible.
π/4 + 1 ≠ π/2 or 3π/2
...
It is a false statement. cos(π/4 + 1) is about –0.213. There is no variable to solve for
cos(a-b)=(cosa)(cosb)-(sina)(sinb)
pi/4=180/4=45 degrees
1=0 degrees
(cos45)(cos0)-(sin45)(sin0)
(sqrt2/2)(1)-(sqrt2/2)(0)
sqrt2/2
cos((π/4) + 1)= cos((π/4) = 45 degrees
you can't. cos (0.25pi + 1 ) is about -.2